Question

${lim}_{x\to 0}\frac{tan2x-sin2x}{x^3}$

Answer 1

${lim}_{x\to 0}\frac{tan2x-sin2x}{x^3}$

$={lim}_{x\to 0}\frac{\frac{sin2x}{cos2x}-sin2x}{x^3}={lim}_{x\to 0}\frac{sin2x(\frac{1}{cos2x}-1)}{x^3}={lim}_{x\to 0}\frac{sin2x(1-cos2x)}{x^{3}cos2x}={lim}_{x\to 0}\frac{sin2x\left(2si{n}^{2}x\right)}{x^{3}cos2x}$

$=\frac{\left({lim}_{x\to 0}\frac{sin2x}{x}\right)\left({lim}_{x\to 0}\frac{2si{n}^{2}x}{x^2}\right)}{\left({lim}_{x\to 0}cos2x\right)}=\frac{({lim}_{x\to 0}\frac{sin2x}{2x}\times 2)\left(2{lim}_{x\to 0}{\left(\frac{sinx}{x}\right)}^2\right)}{{lim}_{x\to 0}cos2x}=\frac{(2\times 1)(2\times 1)}{1}=4$